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When is a surjective polynomial map proper?
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Suppose $F: mathbbC^n to mathbbC^n$ is a surjective function such that $F$ is defined by a polynomial in each coordinate, i.e.
$$ F = (f_1,ldots,f_n) quad f_i in mathbbC[x_1,ldots,x_n] : forall i leq n $$
I know that for $n = 1$, $F$ is simply a polynomial function in one variable, and therefore the fact that $F$ is surjective implies $F$ is proper. Furthermore I know that for $n > 1$, each $f_i$ is not proper as a map $f_i: mathbbC^n to mathbb C$. My question is, under what circumstances is the map $F$ a proper map in general? Is this dependent or independent of whether $F$ is a submersion?
differential-geometry polynomials
asked Jul 24 at 4:57
W. Garland
264
add a comment |Â
up vote
5
down vote
favorite
Suppose $F: mathbbC^n to mathbbC^n$ is a surjective function such that $F$ is defined by a polynomial in each coordinate, i.e.
$$ F = (f_1,ldots,f_n) quad f_i in mathbbC[x_1,ldots,x_n] : forall i leq n $$
I know that for $n = 1$, $F$ is simply a polynomial function in one variable, and therefore the fact that $F$ is surjective implies $F$ is proper. Furthermore I know that for $n > 1$, each $f_i$ is not proper as a map $f_i: mathbbC^n to mathbb C$. My question is, under what circumstances is the map $F$ a proper map in general? Is this dependent or independent of whether $F$ is a submersion?
differential-geometry polynomials
asked Jul 24 at 4:57
W. Garland
264
3
Since positive-dimensional subvarieties of $Bbb C^n$ can never be compact, the only time the map can be proper is if all the fibers are $0$-dimensional.
– Ted Shifrin
Jul 24 at 5:08
Would every fiber being $0$-dimensional be sufficient to show that $F$ is proper?
– W. Garland
Jul 24 at 5:19
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Suppose $F: mathbbC^n to mathbbC^n$ is a surjective function such that $F$ is defined by a polynomial in each coordinate, i.e.
$$ F = (f_1,ldots,f_n) quad f_i in mathbbC[x_1,ldots,x_n] : forall i leq n $$
I know that for $n = 1$, $F$ is simply a polynomial function in one variable, and therefore the fact that $F$ is surjective implies $F$ is proper. Furthermore I know that for $n > 1$, each $f_i$ is not proper as a map $f_i: mathbbC^n to mathbb C$. My question is, under what circumstances is the map $F$ a proper map in general? Is this dependent or independent of whether $F$ is a submersion?
differential-geometry polynomials
asked Jul 24 at 4:57
W. Garland
264
Suppose $F: mathbbC^n to mathbbC^n$ is a surjective function such that $F$ is defined by a polynomial in each coordinate, i.e.
$$ F = (f_1,ldots,f_n) quad f_i in mathbbC[x_1,ldots,x_n] : forall i leq n $$
I know that for $n = 1$, $F$ is simply a polynomial function in one variable, and therefore the fact that $F$ is surjective implies $F$ is proper. Furthermore I know that for $n > 1$, each $f_i$ is not proper as a map $f_i: mathbbC^n to mathbb C$. My question is, under what circumstances is the map $F$ a proper map in general? Is this dependent or independent of whether $F$ is a submersion?
differential-geometry polynomials
asked Jul 24 at 4:57
W. Garland
264
edited Jul 24 at 14:37
asked Jul 24 at 4:57
W. Garland
264
asked Jul 24 at 4:57
W. Garland
264
asked Jul 24 at 4:57
W. Garland
264
264
3
Since positive-dimensional subvarieties of $Bbb C^n$ can never be compact, the only time the map can be proper is if all the fibers are $0$-dimensional.
– Ted Shifrin
Jul 24 at 5:08
Would every fiber being $0$-dimensional be sufficient to show that $F$ is proper?
– W. Garland
Jul 24 at 5:19
add a comment |Â
3
Since positive-dimensional subvarieties of $Bbb C^n$ can never be compact, the only time the map can be proper is if all the fibers are $0$-dimensional.
– Ted Shifrin
Jul 24 at 5:08
Would every fiber being $0$-dimensional be sufficient to show that $F$ is proper?
– W. Garland
Jul 24 at 5:19
3
3
Since positive-dimensional subvarieties of $Bbb C^n$ can never be compact, the only time the map can be proper is if all the fibers are $0$-dimensional.
– Ted Shifrin
Jul 24 at 5:08
Since positive-dimensional subvarieties of $Bbb C^n$ can never be compact, the only time the map can be proper is if all the fibers are $0$-dimensional.
– Ted Shifrin
Jul 24 at 5:08
Would every fiber being $0$-dimensional be sufficient to show that $F$ is proper?
– W. Garland
Jul 24 at 5:19
Would every fiber being $0$-dimensional be sufficient to show that $F$ is proper?
– W. Garland
Jul 24 at 5:19
add a comment |Â
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3
Since positive-dimensional subvarieties of $Bbb C^n$ can never be compact, the only time the map can be proper is if all the fibers are $0$-dimensional.
– Ted Shifrin
Jul 24 at 5:08
Would every fiber being $0$-dimensional be sufficient to show that $F$ is proper?
– W. Garland
Jul 24 at 5:19